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D01-24

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Re: D01-24 [#permalink]

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New post 29 Dec 2015, 01:08
Dear Buffaloboy,

Please note that 24 kms from point p are not covered in 01 hour. See the reasoning below:

- Difference in time between two buses 1 hour

- Total time required to cover the distance: 04 hours

Y Y Y Y Y Y X X X X (Here X X X X reflects the distance covered in 01 hour by bus leaving earlier; Please note its just for understanding the reasoning)

- Not pending distance to be covered is - Y Y Y Y Y Y -

- As the speed is constant for both buses, each bus would take similar time to cover half distance (point where they meet; 3 Ys in our example)

- Now remember from question stem that 04 hours required to travel the distance

- Hence total time for buses would be 1.5 (half distance of Y Y Y Y Y Y by one bus) + 1.5 Y Y Y Y Y Y (half distance of Y Y Y Y Y Y by other bus) + 1 hour (extra one)

From there on, I think explanation is self explanatory.

Hope it helps!

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Re D01-24 [#permalink]

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I think this is a poor-quality question and I agree with explanation. I wasted a minute or so trying to figure out if there was a trick I wasn't getting but then realized the problem is worded incorrectly. If they already met at P and they do a "return" trip, they will not be meeting again..... It should say, if they do "the same trip the next day but...."

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Re D01-24 [#permalink]

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New post 18 Mar 2016, 21:07
I think this is a poor-quality question. It might be useful to state that they continue on their journey towards N and M respectively. "They meet at point P" made it seem like they stopped at point P and then started on the return journey from P.

I might have wrongly deduced that, but you may want have another look.

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Re D01-24 [#permalink]

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New post 19 Apr 2016, 23:23
I think this is a high-quality question and I agree with explanation.

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Re: D01-24 [#permalink]

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New post 16 Jun 2016, 14:05
qw1981 wrote:
its given that 1 bus starts 36 minutes late and the other 24 minutes early which means effectively 1 bus had head start of 1 hr. Since they are traveling at same speed and they meet 24 miles from where they met yesterday then the bus which had head start of 1 hr traveled 24 miles in 1 hr. Thus its speed is 24 miles. The speed of 2nd bus is also 24 (since both bus travels at constant speed)..
Thus both bus travels for 4 hrs hence distance traveled by 1 bus =24*4

Total distance= 2*24*4= 192.. Answer


Can someone enlighten me on the above explanation?
If a bus travels 24miles/h; the total time is 4 hours; total distance is 24*4= 96.
Why are we supposed to multiply 96*2? :?:

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Re D01-24 [#permalink]

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New post 27 Jun 2016, 02:20
I think this is a high-quality question and I agree with explanation.

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D01-24 [#permalink]

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New post 08 Jul 2016, 07:00
I think this is a high quality question .

To solve it I used a straight approach, converting a S*T=D problem into a R*T=W problem.
The distance between the 2 cities becomes the work => W=MN
We know that the 2 bus are driving toward each other a constant speed v therefore in 1h their summed speed is gonna be 2v
We know that the time is 2h
r-------t------w
2v-----2------MN => basically we are answering the question: " how long will it take to a constant rate of 2v km/h to close a gap of MN km?"

so we can deduce that 4v=MN => v=MN/4

r---------t------ w
MN/4----2------.5MN => we know the speed and the time => we can calculate the work done = distance covered (by one bus) => .5MN=P

Then if one bus leaves 36 min earlier than scheduled, while the other leaves 24 min later than scheduled therefore for 1 h one of the two bus moves while the other doesn't

r-------- t-------w
MN/4----1------MN/4 => in this hour one bus covers MN/4 therefore when the second bus depart so that the two bus are driving toward each other there still 3MN/4 to be covered which is gonna be the work.

r---------t------ w
MN/2----1.5----3MN/4 => speed is MN/2 because the rate at which the gap (3MN/4) is closing is 2v

r------------t---------w
MN/4------1.5-------3MN/8 the new meeting point P' will be 3MN/8


in conclusion MN/2= (3MN/8) +24 => MN=24*8= 192

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Re D01-24 [#permalink]

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New post 31 Aug 2016, 11:10
I think this is a poor-quality question. The question is confusing

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Re D01-24 [#permalink]

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New post 05 Sep 2016, 06:19
I think this is a poor-quality question and I agree with explanation. The question needs to be phrased better. It is extremely confusing and such confusion of deciphering how the scenario unfolds rarely happens on the actual GMAT. It is unclear from the question that both buses actually make the trip to the other cities. The 'After driving for 2 hours they meet at point P' leads to believe that both buses make the trip just till point P. This confusion is further amplified by the next statement that they make the return trip and then meet 24 kms away from point P. It took me 2 minutes just to make sense of the question. How can they meet if they're making the return trip from point P towards their respective cities. Their paths do not cross!

Solid question otherwise. Very challenging.

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Re: D01-24 [#permalink]

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New post 17 Dec 2016, 10:50
|-------|-------|-------|-------|

M--------------------------------N

Note that each bus travel at the same speed. Therefore they travel equal distance in equal time.

Each bus travels halfway in 2 hours. Therefore each bus travels travel 1/4th in 1 hour

On 2nd day M had a 1 hour head start. Therefore, M travel for 1/4 the distance before N started.

After 1 hour

|-------|-------|-------|-------|

---------M-----------------------N

Now they will travel equal distances before they meet.

|-------|-------|---|--|-------|

--------------------MN--------
They meet halfway between 1/2 and 3/4 i.e. 5/8.

That is 1/8 to the right of 1/2

|-------|-------|24|---|-------|

And 1/8 is equal to 24 miles.

Therefore the entire distance is 8 x 24= 192.

The problem is that simple. Ironically, I started with a RTD chart and guessed after 3 minutes.

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Re: D01-24 [#permalink]

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New post 26 Dec 2016, 19:41
buffaloboy wrote:
I think this is a high-quality question and I don't agree with the explanation. In second scenario, when one bus is delayed for 24 minutes and another bus leaves 36 minutes , it is clear that one bus has traveled for one hour extra before they meet. Had they traveled for same time they would meet at P but they meet 24 miles from point P. This means this distance is traveled in one hour by one of the bus that has traveled one hour more than other by the time they meet. So speed of bus would be 24 miles per hour. since it takes 4 hours so distance would be 96 miles.


I thought the same at first until drawing it out. The above is good up until the "this means this distance is traveled in one hour..." part.

From the question, it is clear that the total travel time is 4 hours, and on the second day one of the buses gets an hour head start (ie. |t1-t2|= 1 hour). On day 2, after traveling alone for 1 hour, 3 hours are left to go. Since the buses travel at the same speed, they have to meet in the middle, or 1.5 hours from now. Hence, when they meet, one bus has been traveling for 1+1.5=2.5 hours and the other has been traveling 1.5 hours.

Going back to your statement, "this means the distance is traveled in" 30 minutes. Think of it this way. Point P is 2 hours away. Point P + 24 miles is 2.5 hours away, per the above. Hence, 24 miles is traveled in 30 minutes. 24*(4 hours)/(30 minutes) = 24*8=192 miles.

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Re: D01-24 [#permalink]

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New post 26 Feb 2017, 11:19
Apply r-t-d table to visualize in a simple manner.
Given in the ques.
r t d
Bus1 x 2 2x
Bus2 x 2 2x

Now, as per the updated info
r t d
Bus1 x 1+t x(1+t) ...Since we know one bus had a head start of 24+36 min = 60 min = 1 hr
Bus2 x t xt .....t is taken because we know that speed of both buses is constant. So, both will travel for same time before meeting.

Now, we know, at meeting point, For Bus 1: 2x+24= x(1+t)..................(1)
& at meeting point for Bus 2: 2x-24= xt...................(2)
Solving (1) & (2). We get, x=48. Since we know from Ist table. Total distance f=b/w two placed is 4x => 4*48= 192.

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Re: D01-24 [#permalink]

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New post 16 Apr 2017, 15:01
Bunuel wrote:

A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\).

The following day the buses do the return trip at the same constant speed.

One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point \(P\), what is the distance between the two cities?


This question confused me. At first I thought it was impossible for the busses to meet at P on day 2. The question never states that the busses reach their destination on day 1, so I thought they stayed at point P on day 1. The reader has to make the assumption that the busses PASS each other on point P and reach the opposite directions

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Re D01-24 [#permalink]

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New post 22 Sep 2017, 23:30
I think this is a high-quality question and I don't agree with the explanation. What is wrong with the following approach?

In the second scenario, when one bus is delayed for 24 minutes and another bus leaves 36 minutes, it is clear that one bus has traveled for one hour extra before they meet. Had they traveled for same time they would meet at P but they meet 24 miles from point P. This means this distance is traveled in one hour by one of the buses that have traveled one hour more than other by the time they meet. So the speed of bus would be 24 miles per hour. Since it takes 4 hours so distance would be 96 miles.

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Re: D01-24 [#permalink]

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New post 23 Sep 2017, 04:06
Subham10 wrote:
I think this is a high-quality question and I don't agree with the explanation. What is wrong with the following approach?

In the second scenario, when one bus is delayed for 24 minutes and another bus leaves 36 minutes, it is clear that one bus has traveled for one hour extra before they meet. Had they traveled for same time they would meet at P but they meet 24 miles from point P. This means this distance is traveled in one hour by one of the buses that have traveled one hour more than other by the time they meet. So the speed of bus would be 24 miles per hour. Since it takes 4 hours so distance would be 96 miles.


If you don't understand the solution please check other discussion here: https://gmatclub.com/forum/a-bus-from-c ... 86478.html

Hope it helps.
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Re: D01-24 [#permalink]

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New post 19 Oct 2017, 06:20
My alternative solution:
Let's rid out of the delayed bus and continue with just one bus that left earlier. What does it give us?
The bus can handle distance between M and N for 4 hours. P is the midpoint between the cities.
According to the question stem, one bus left 36 min earlier and another delayed for 24 min. It is the same as saying that early-bird bus left 60 min earlier (24+36 = 60).
Let's say that he must have been living at 8 am, but he actually left at 7 am.
So, at 10 am.
Planned time: 2 hours
Planned distance: MP
Fact time: 3 hours
Fact distance: MP+48 miles (48 miles = 24 miles*2. Why times 2? Since we considering the speed of the second bus as 0, therefore, nobody is driving toward us and we have to drive distance of the second bus by ourselves, in addition to our own distance)
Therefore, the speed is 48 m/h. And the distance is 48*4 = 192.

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Re: D01-24 [#permalink]

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New post 21 Oct 2017, 23:34
Mind = Blown ! Brilliant explanation Bunuel

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Re: D01-24   [#permalink] 21 Oct 2017, 23:34

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